Suppose $F_{\mu\nu}$ and $F^{\mu\nu}$ are defined as follows:
$$F_{\mu\nu} = \begin{bmatrix} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_z & 0\end{bmatrix}, F^{\mu\nu} = \begin{bmatrix} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_z & 0\end{bmatrix}$$
We define $L_{Maxwell}$ to be:
$$L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}$$ $$=-\frac{1}{4}(-2E_x^2-2E_y^2-2E_z^2+2B_x^2+2B_y^2+2B_z^2)$$ $$=\frac{1}{2}(E_x^2 + E_y^2 + E_z^2 - B_x^2 - B_y^2 - B_z^2)$$
Extremizing for $L_{\vec A + \vec \epsilon} - L_{\vec A}$, we obtain the Euler-Lagrange equation:
$$ \color{blue}{E_x + E_y + E_z - B_x - B_y - B_z = 0}$$
Would this be the correct form of the Euler-Lagrange equation for L?
I can't exactly see how the above solution is equivalent to Maxwell's equations in free space?
$$\nabla \cdot \vec E = 0$$ $$\nabla \cdot \vec B = 0$$ $$\nabla \times \vec E = -\frac{\partial\vec B}{\partial t}$$ $$\nabla \times \vec B = \frac{\partial \vec E}{\partial t}$$
The Euler-Lagrange condition splits into n independent simultaneous equations per each field component.
$$d_t(\frac{dL}{dA_{0,t}})+d_x(\frac{dL}{dA_{0,x}})+d_y(\frac{dL}{dA_{0,y}})+d_z(\frac{dL}{dA_{0,z}}) - \frac{dL}{dA_0}=0\\ d_t(\frac{dL}{dA_{1,t}})+d_x(\frac{dL}{dA_{1,x}})+d_y(\frac{dL}{dA_{1,y}})+d_z(\frac{dL}{dA_{1,z}})- \frac{dL}{dA_1}=0\\ d_t(\frac{dL}{dA_{2,t}})+d_x(\frac{dL}{dA_{2,x}})+d_y(\frac{dL}{dA_{2,y}})+d_z(\frac{dL}{dA_{2,z}})- \frac{dL}{dA_2}=0\\d_t(\frac{dL}{dA_{3,t}})+d_x(\frac{dL}{dA_{3,x}})+d_y(\frac{dL}{dA_{3,y}})+d_z(\frac{dL}{dA_{3,z}})- \frac{dL}{dA_3}=0\\$$
For L defined as
$$L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} $$ $$= \frac{1}{2}[(E_x)^2+(E_y)^2+(E_z)^2-(B_x)^2-(B_y)^2-(B_x)^2]$$
$$=\frac{1}{2}[(\frac{dA_0}{dx})^2+(\frac{dA_0}{dy})^2+(\frac{dA_0}{dz})^2 -(\frac{dA_3}{dy}-\frac{dA_2}{dz})^2 - (\frac{dA_1}{dz} - \frac{dA_3}{dx})^2 - (\frac{dA_2}{dx} - \frac{dA_1}{dy})^2]\\$$
E-L equation 1 gives:
$$d_t(0) + d_x(\frac{dA_0}{dx}) + d_y(\frac{dA_0}{dy}) + d_z(\frac{dA_0}{dz}) - (0)= 0\\ \frac{dE_x}{dx} + \frac{dE_y}{dy} + \frac{dE_z}{dz} = 0\\ \color{red}{\nabla \cdot \vec E = 0}\\$$
which is Maxwell's 1st equation in free space.
E-L equation 2 gives:
$$d_t(0) + d_x(0) + d_y(\frac{dA_2}{dx} -\frac{dA_1}{dy}) + d_z(\frac{dA_3}{dx} - \frac{dA_1}{dz}) - (0)= 0\\ \frac{dB_z}{dy} - \frac {dB_y}{dz} = 0\\$$
E-L equation 3 gives: $$\frac{dB_x}{dz} - \frac {dB_z}{dx} = 0\\$$
E-L equation 4 gives: $$\frac{dB_y}{dx} - \frac {dB_x}{dy} = 0\\$$
which three together gives
$$\color{red}{\nabla \times \vec B = 0}$$
which is Maxwell 4th equation for when $\vec j=0$ and $\frac{d \vec E}{dt} = 0$.
Charge source density $\rho$ and charge movement $\vec j$ enters via
$$L_{QED}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + i\bar \Psi \gamma^\mu\frac{d\Psi}{d\mu}- m\bar \Psi \Psi - q_e\bar \Psi\gamma^\mu A_\mu\Psi$$
in which case
$$\frac{dL}{dA_\mu} = q_e\bar \Psi\gamma^\mu \Psi$$
and
$$\color{red}{\nabla \cdot \vec E = q_e\bar \Psi\gamma^0 \Psi}\\$$ $$\color{red}{\nabla \times \vec B = \begin{pmatrix} q_e\bar \Psi\gamma^1 \Psi \\ q_e\bar \Psi\gamma^2 \Psi \\ q_e\bar \Psi\gamma^3 \Psi\end{pmatrix}}$$